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COMPUTING THE FROBENIUS NUMBER
Abdallah Badra
To cite this version:
Abdallah Badra. COMPUTING THE FROBENIUS NUMBER. International Journal of Pure and Applied Mathematics, Academic Publishing Ltd, 2010, 60 (1), pp.89-105. �hal-00477408�
COMPUTING THE FROBENIUS NUMBER
Abdallah Badra
Laboratoire de Math´ematiques, Universit´e Blaise Pascal, Les C´ezeaux,
63177 Aubi`ere Cedex, France
E-mail : abdallah.badra@math.univ-bpclermont.fr
Abstract. The Frobenius number g(A) of a finite subset A ⊂ IN such that gcd(A) = 1 is the largest integer which cannot be expressed as P
a∈Aaxa with non-negative integers xa. We present
an algorithm for the computation of g(A). Without loss of general-ity we suppose that there exist a, b ∈ A such that gcd(a, b) = 1. We give a formula for g(A) in the particular case that for all c, d ∈ A, c + d can be written in the form c + d = xa + yb with x, y ≥ 0 (e.g. c + d > ab − a − b). Using Euler polynomials we give a formula for g(A) in the case that A = {a, b, c}.
AMS subj. Classification: 11D04
Key Words: Frobenius number, linear diophantine equation.
1. Introduction and Statement of Results
Throughout this paper, small letters denote integers. We will set IN = {1, 2, 3 . . . } and IN0 = IN ∪ {0}. We consider a finite subset
A ⊂ IN such that gcd(A) = 1. We define S(A) the additive semi-group of IN0 generated by A i.e. S(A) = {Pa∈Axaa | xa ∈ IN0}.
The Frobenius number g(A) is defined as the largest integer which does not belong to S(A). We are interested in computing g(A) and equivalently f (A) = g(A) +P
a∈Aa the largest integer which
can-not be expressed as P
a∈Aaxa with positive integers xa. It is well
and only if 1 ∈ A.
For card(A) ≥ 3, no general formula for g(A) is known, except in particular cases, see [6] and [8]. Algorithms are developed in [4] and [7] in the case A = {a, b, c}.
Without loss of generality, we can consider only sets A containing two coprime integers a, b (see section 2). We fix two integers a > 1 and b > 1 such that gcd(a, b) = 1, and c = (c1, . . . , cn) ∈ INn. Let
A = {a, b, c1, . . . , cn}.
For all t = (t1. . . , tn) ∈ INn0, we set t.c =
Pn
1tici.
For every m ∈ ∠Z, there exist unique integers ˆm and ˇm such that m = ˆma − ˇmb with 0 ≤ ˆm < b. We write t.c = b(t)a − a(t)b where b(t) =t.c and a(t) =∧ t.c.∨
A finite subset T of INn0 is said to be appropriate if S(A) = [
t∈T
(S(a, b) + t.c). (1)
It is said to be pruned if 0 ∈ T and for t ∈ T \ {0}, we have b(t) > 0 and a(t) > 0.
Let T be an appropriate and pruned subset of INn0. We can write T = {0 = t0, t1, . . . , tl} with 0 = b(t0) < b(t1) ≤ b(t2) ≤ · · · ≤
b(tl) < b. We set bi = b(ti), ai = a(ti) for 0 ≤ i ≤ l, bl+1 = b and
al+1 = a.
We obtain the sequences
0 = b0 < b1 ≤ b2 ≤ · · · ≤ bl < bl+1 = b
and
0 = a0 < a1, a2, . . . , al < al+1 = a.
For 0 ≤ i ≤ l, we set mi = max{aj | 0 ≤ j ≤ i} and
gi = (bi+1− 1)a − (mi+ 1)b.
Theorem 1.
g(A) = max{gi | 0 ≤ i ≤ l}.
(2)
A subset M ⊂ IN0 is said to be trimmed if for all m ∈ M \
{0}, ˇm > 0 and for all m, d ∈ M, ˆ
m < ˆd ⇔ ˇm < ˇd. (3)
We show that there exists an appropriate set T such that T.c = {t.c | t ∈ T } is trimmed.
Theorem 2. Let T be an appropriate set. If T.c is trimmed then g(A) = max{a(bi+1− bi) + ti.c | 0 ≤ i ≤ l} − (a + b)
(4)
where t0 = 0.
We give an algorithm for computing g(A). We use Theorem 1 if we start with an arbitrary appropriate set T . And we use Theorem 2 if we start with an appropriate set T such that T.c is trimmed. In a particular case we derive a formula for g(A) from Theorem 2: Theorem 3. If {c1, . . . , cn} is trimmed and for all 1 ≤ i, j ≤ n,we
have ci+ cj ∈ S(a, b) (e.g. ci+ cj > ab − a − b) then
g(A) = max{a(bi+1− bi) + ci | 0 ≤ i ≤ n} − (a + b)
(5)
where bn+1 = b, b0 = 0, c0 = 0 and for all 1 ≤ i ≤ n, bi = ˆci.
In the case n = 1, i.e. A = {a, b, c}, we take c 6∈ S(a, b) and we write c = au0− bv0 with 0 < u0 < b and 0 < v0 < a.
By successive Euclidean divisions we get: (
a = u−1 = p0v0+ u1, . . . , un−1 = pnvn+ un+1, . . .
b = v−1 = q0u0+ v1, . . . , vn−1= qnun+ vn+1, . . .
The triplet (a, b, c) is said to be of level n = n(a, b, c) if p0 = q0, p1 = q1, · · · , pn−1= qn−1
and (pn6= qn or vn+1 = 0).
Let (a, b, c) a triplet of level n. For all 0 ≤ i ≤ n, we set wi+1 =
ui−1− qivi so for i ≤ n, wi = ui and wn+1 = (pn− qn)vn+ un+1.
We denote by L(a, b, c) = (L0, L1) = ( (bvn, bwn+1) if n is even (avn, awn+1) if n is odd , l(a, b, c) = (l0, l1) = ( (awn, avn+1) if n is even (bwn, bvn+1) if n is odd .
For an integer n we denote by k(n) = k = bn2c and h(n) = h = bn+1
2 c where bxc is the greatest integer such that bxc ≤ x.
Theorem 4.
f (a, b, c) = aw2h+ bw2k+1− min{bv2h, av2k+1}
= l0+ L1− min{l1, L0}.
(7)
In particular, when pn = qn and vn+1 = 0,
f (a, b, c) = (
a gcd(b, c) + lcm(b, c) if n is even b gcd(a, c) + lcm(a, c) if n is odd. (8)
Theorem 5. Let (x−1, x0, . . . , xl+1), (y−1, y0, . . . , ym+1), (d0, d1, . . . , dl)
and (e0, e1, . . . , em) be sequences such that the following conditions
( x−1 = d0x0 − x1, . . . , xl−1 = dlxl− xl+1, y−1 = e0y0− y1, . . . , ym−1 = emym− ym+1, (9) ( a = x−1 > x0 > x1 > · · · > xl > xl+1 = 0, b = y−1 > y0 > y1 > · · · > ym > ym+1 = 0 (10)
hold. Let n be an integer such that ( d0 = e0, d1 = e1, · · · , dn−1 = en−1 and (dn 6= en or xn+1 = 0). (11) Then, f (a, b, c) = ayn+1+ b(xn−1− enxn) + max{a(yn− yn+1), bxn}. (12)
As a consequence of Theorem 5 we prove the following theorem see [7].
Theorem 6(Rodseth). Let Ri polynomials defined by induction
as follows: R−1 = 0, R0 = 1 and for all i > 0,
Ri+1= eiRi− Ri−1.
(13)
Let n be the unique integer such that yn+1 Rn+1 ≤ c a < yi Ri for all 0 ≤ i ≤ n. (14) Then, f (a, b, c) = cRn+1+ ayn− min{ayn+1, cRn}. (15)
2. Reduction to the case gcd(a, b) = 1
Let A = {b, c1, . . . , cn} be a subset of IN such that gcd(A) = 1.
We recall the following classical result: It is always possible to choose integers x, x1. . . , xn such that 1 = xb + x1c1+ · · · + xncn.
Then, for k an integer sufficiently large, we get
Hence, g(A) = g(A ∪ {a}) and gcd(a, b) = 1. In the particular case that
A = {a, b, c1, . . . , cn} such that gcd(a, b) = gcd(a, b, c1, . . . , cn−1) =
d ≥ 1 we observe that gcd(ad,bd) = 1. Therefore, to compute g(A) we can use Brauer’s formula
f (A) = df (a d, b d, c1 d, . . . , cn−1 d , cn), (16) see [3].
From now on we suppose that gcd(a, b) = 1. 3. Case n ≥ 1 We need some remarks.
R. If m ≥ 0 then a > ˇm.
Proof. Since m = ˆma − ˇmb ≥ 0 we have ba > ˆma ≥ ˇmb hence a > ˇm 2
R2. Let m = xa − yb. Then,
0 ≤ x < b ⇔ ˆm = x ⇔ ˇm = y and
−b ≤ x < 0 ⇔ ˆm = x + b ⇔ ˇm = y + a.
Proof. To prove the second claim we write m = xa − yb = (x + b)a − (y + a)b and we use the uniqueness of ˆm and ˇm 2 R. Let m = xa − yb. Then, there exists a unique integer p ∈ ∠Z such that x = pb + ˆm and y = pa + ˇm.
Proof. We write m = xa−yb = ˆma− ˇmb thus (x− ˆm)a = (y − ˇm)b. Since gcd(a, b) = 1, p = x− ˆbm = y− ˇam is an integer 2
R4. We have m ∈ S(a, b) if and only if ˇm ≤ 0.
Proof. Clearly ˇm ≤ 0 ⇒ m = ˆma − ˇmb ∈ S(a, b). Conversely, if m ∈ S(a, b) then m = xa + yb with x ≥ 0 and y ≥ 0. By R3 there exists p ∈ ∠Z such that x = pb + ˆm ≥ 0 thus p ≥ 0. We also have
−y = pa + ˇm ≤ 0 and so ˇm = −y − pa ≤ 0 2 R5. For d ≥ 0 we set G(d) = S(a, b)S(S(a, b) + d). Then, we have
m 6∈ G(d) ⇔ ˇm > 0 and ( ˆm < ˆd or ˇm > ˇd).
Proof. Let n = m − d. Since d ≥ 0, R1 shows that a > ˇd. Hence, ˇ
m − ˇd + a > ˇm. Moreover, we have n = ( ˆm − ˆd)a − ( ˇm − ˇd)b and −b < ˆm − ˆd < b. It follows from R2 that ˇn = ˇm − ˇd if ˆm − ˆd ≥ 0 and ˇn = ˇm − ˇd + a > ˇm if ˆm − ˆd < 0. We deduce that
ˇ m > 0 and ( ˆm < ˆd or ˇm > ˇd) ⇔ ˇ m > 0 and ( ˆm − ˆd ≥ 0 ⇒ ˇm − ˇd > 0) ⇔ ˇ m > 0 and ˇn > 0 ⇔
m 6∈ S(a, b) and n = m − d 6∈ S(a, b) ⇔
m 6∈ G(d)2 As a consequence of R5 we obtain R6. If T is appropriate then
m 6∈ S(A) ⇔ ˇm > 0 and ∀t ∈ T, ˆm < b(t) or ˇm > a(t). Proof of Theorem 1. For 0 ≤ i ≤ l, we set
F (i) = {m ∈ ∠Z | ˆm < bi+1 and ˇm > mi}.
Step 1. ∠Z \ S(A) = l [ i=0 F (i). (17)
Let m ∈ F (i). For every 0 ≤ j ≤ l, ˆm < bi+1 ≤ bj if j ≥ i + 1 and
ˇ
m > mi ≥ aj if j ≤ i. Hence, R6 shows that m 6∈ S(A).
Conversely, let m 6∈ S(A) then ˇm > 0 by R4. Since b0 = 0 ≤ ˆm <
b = bl+1 and 0 = b0 < b1 ≤ b1 ≤ · · · ≤ bl ≤ bl+1 = b, there exists
0 ≤ j ≤ l such that ˆm < bj+1. We put i = min{j | ˆm < bj+1}. We
thus get for 0 ≤ j ≤ i, bj ≤ bi ≤ ˆm < bi+1 and ˇm > aj by R6.
Hence, ˇm > mi. We conclude that m ∈ F (i).
Step 2. For 0 ≤ i ≤ l, gi = max F (i).
Since 0 ≤ bi+1− 1 < b, we have ˆgi = bi+1− 1 and ˇgi = mi+ 1 > mi.
We thus get gi ∈ F (i). Moreover, for all m ∈ F (i), m = ˆma − ˇmb ≤
(bi+1− 1)a − (mi + 1)b then gi = max F (i)2
In particular, when T = {0} i.e. when all ci ∈ S(a, b) we have
b0 = 0 < b1 = b, a0 = 0 and m0 = 0. Therefore, g(A) = g0 =
(b1− 1)a − (m0+ 1)b = ab − a − b = g(a, b).
Proof of Theorem 2. If T .c is trimmed then 0 = b0 < b1 < b1 <
· · · < bl < bl+1 = b and thus 0 = a0 < a1 < a1 < · · · < al < al+1 =
a. In particular, T is pruned and for all 0 ≤ i ≤ l, mi = ai. We can
write gi = a(bi+1−bi)+abi−bai−(a+b) = a(bi+1−bi)+ti.c−(a+b)
2
Algorithm.
1. For every i, we choose λi > 0 such that λici ∈ S(a, b).
Num-bers λi exist. Indeed, it is sufficient to take λi > g(a,b)
ci =
ab−a−b
ci . The following set
U = {t = (t1, . . . , tn) ∈ INn0 | ti < λi, 1 ≤ i ≤ n}
is thus appropriate.
We remove from U all elements t 6= 0 such that b(t) = 0 or a(t) ≤ 0. The set T of all remainding elements is pruned and still appropriate.
way that 0 = b(t0) < b(t1) ≤ b(t2) ≤ · · · ≤ b(tl). We compute
mi = max{aj | 0 ≤ j ≤ i}, gi = (bi+1− 1)a − (mi + 1)b and
g(A) = max{gi | 0 ≤ i ≤ l} where bi = b(ti), ai = a(ti) for
0 ≤ i ≤ l, bl+1 = b and al+1= a.
2. The algorithm can be modified as follows: For every 1 ≤ i ≤ n, we start removing from A all elements cj such that ˆci ≤ ˆcj
and ˇcj ≤ ˇci. We choose an appropriate and pruned set T =
{0 = t0, t1, . . . , tl}. We can suppose that 0 = b(t0) < b(t1) ≤
b(t2) ≤ · · · ≤ b(tl) < b. For all 0 ≤ i ≤ l, we remove from T
all tj such that (b(ti) ≤ b(tj) and a(tj) ≤ a(ti)). Considering
the set of the remainding elements we can suppose that T.c is trimmed. The Frobenius number can therefore be computed using Theorem 2.
Proof.
1. For all t ∈ INn0, we have G(t.c) = S(a, b)S(S(a, b) + t.c) ⊂ S(A) thus S
t∈UG(t.c) ⊂ S(A). Conversely, let m ∈ S(A),
then m = xa + yb +Pn
i=1xici with x ≥ 0, y ≥ 0 and xi ≥ 0.
By Euclidean division we write xi = qiλi + ti. We thus get
t = (t1, . . . , tn) ∈ U and m = xa + yb +
Pn
i=1qiλici + t.c.
Since λici ∈ S(a, b), we have m ∈ G(t.c). The equality
S(A) = [
t∈U
G(t.c) (18)
follows. Therefore, U is appropriate.
By construction T is pruned and it is still appropriate. Indeed, for t 6= 0, t.c 6∈ S(a, b)} if and only if a(t) > 0 by R4. Therefore,
T = {t ∈ U | t = 0 or t.c 6∈ S(a, b)} (19)
and S(A) = [ t∈T G(t.c). (20) 2. If ˆci ≤ ˆcj and ˇcj ≤ ˇci then cj = ci+ (ˆcj − ˆci)a + (ˇci− ˇcj)b ∈
S(ci, a, b). Therefore, g(A) = g(A \ {cj}). If b(ti) ≤ b(tj)
and a(tj) ≤ a(ti) then tj.c = ti.c + (b(tj) − b(ti))a + (a(ti) −
a(tj))b ∈ G(ti.c) thus G(tj.c) ⊂ G(ti.c). We see that (20) is
not altered by removing tj from T 2
Proof of Theorem 3. Since 2ci ∈ S(a, b) for all 1 ≤ i ≤ n,
U = {t = (t1, . . . , tn) ∈ INn0 | 0 ≤ ti ≤ 1, 1 ≤ i ≤ n} is appropriate.
Furthermore, for t = (t1, . . . , tn) ∈ U , if t.c 6∈ S(a, b) then there
exists ci and y ∈ S(a, b) such that t.c = ci + y ∈ G(ci).
There-fore, the set T = {0 = t0, t1, . . . , tn} where ti = (0, . . . , ti, . . . , 0)
and ti = 1 is also appropriate. Moreover, T .c = {0, c1, . . . , cn} is
trimmed and b(ti) = ˆci for all 1 ≤ i ≤ n 2
4. Case n = 1
Let E be a totally ordered set and x ∈ E. We call successor of x in E, and we denote x+, the smallest element of E (if there exists
any) such that x < x+.
For an appropriate set T , we put b(T ) = {b(t) | t ∈ T } and B = b(T ) ∪ {b}. We equip B with the natural order ≤.
To apply Theorem 2, in the case that n = 1, it is convenient to formulate it as follows:
Theorem 20. Let T be an appropriate set such that T .c is trimmed. Then,
g(A) = max{a(b(t)+− b(t)) + t.c | t ∈ T } − (a + b) (21)
where b(t)+ is the successor of b(t) in B.
Suppose that there exists integers α1, . . . , αqand a partition E1, . . . , Eq
of T such that b(t)+− b(t) = α
i for all t ∈ Ei and for all 1 ≤ i ≤ q.
Then,
g(A) = max{aαi+ βi | 1 ≤ i ≤ q} − (a + b)
(22)
where we put βi = max Ei.c.
To compute such a partition, we are led to introduce what we call Euler order on T . We will use Euler polynomials.
Euler polynomials
Let (q0, q1, . . . , qi, . . . ) be a sequence of positive integers. We define
Euler polynomials Qi by induction as follows: Q−1 = 0, Q0 = 1 and
for i ≥ 0,
Qi+1(q0, . . . , qi) = qiQi(q0, . . . , qi−1) + Qi−1(q0, . . . , qi−2).
(23)
We set Qi+1 = Qi+1(q0, . . . , qi), Q1i = Qi(q1, . . . , qi), Pi+1 = Qi +
Qi+1 and Pi+11 = Q1i + Q1i+1.
We deduce immediately that
Qn+1 = qnQn+ qn−2Qn−2+ . . . + qn−2iQn−2i+ Qn−2i−1
(24)
for 0 ≤ 2i ≤ n. Euler order ≤e
Proposition 1. Every integer t ∈ IN0 can be written uniquely in
the form t = t0Q0+ t1Q1+ · · · + tnQn (25) where min{i | ti > 0} is even , (26)
0 ≤ ti ≤ qi for 0 ≤ i ≤ n
(27) and
ti = qi ⇒ ti−1 = 0 for 1 ≤ i ≤ n.
(28)
Equality (25) is called Euler expansion of t.
Proof. There exists n ∈ IN0 such that t < Qn+1. By successive
Euclidean divisions we can write t = tnQn+ sn with 0 ≤ sn < Qn, sn = tn−1Qn−1+ sn−1 with 0 ≤ sn−1 < Qn−1, .. . s2 = t1Q1+ s1 with 0 ≤ s1 < Q1, s1 = t0Q0 with t0 = s1.
We put i = min{j | tj > 0}. If i = 2e then t = t2eQ2e+t2e+1Q2e+1+
· · · + tnQn is Euler expansion of t. If i = 2e + 1, using (24)
we take t = q0Q0 + · · · + q2eQ2e + (t2e+1 − 1)Q2e+1 + · · · + tnQn
as Euler expansion of t. Conditions (27) and (28) follow from si+1 < Qi+1= Qi−1+qiQi. The uniqueness follows from the
unique-ness of the Euclidean division and the fact that P2e
i=0tiQi < Q2e+1
if and only if there exists 0 ≤ j ≤ e such that t2j < q2j 2
For t, x ∈ IN0, let n ∈ IN0 such that t, x ≤ Qn+1. We consider
Euler expansions of t and x respectively
t = t0Q0+ t1Q1+ · · · + tnQn and x = x0Q0+ x1Q1+ · · · + xnQn.
We define Euler order ≤e as follows: t ≤ex if
(t0, −t1, . . . , (−1)iti, . . . , (−1)ntn) ≤l (x0, −x1, . . . , (−1)ixi, . . . , (−1)nxn)
where ≤l is the lexicographic order on ∠Zn.
Lemma. Let n ∈ IN0. We consider Un = {0, 1, . . . , Qn+1− 1} and
1. We define a partition of Un= F1∪ F2 as follows: F1 = ( {0, 1, . . . , Q2k+1− Q2k− 1} if n = 2k {Q2k+1, . . . , Q2k+2− 1} if n = 2k + 1 F2 = ( {Q2k+1− Q2k, . . . , Q2k+1− 1} if n = 2k {0, 1, . . . , Q2k+1− 1} if n = 2k + 1 .
Then, the successor, in Un, of all t ∈ F1 (resp. t ∈ F2) is t+ =
t + (−1)nQn (resp. t+= t + (−1)n[Qn− Qn+1]). In particular,
if qn= 1 then for all t ∈ F2, t+= t + (−1)n−1Qn−1.
2. We define a partition of Vn = E1 ∪ E2 as follows: E1 =
{0, . . . , Q2h−1 − 1}, E2 = {Q2h−1, . . . , Pn − 1}. Then, the
successor, in Vn, of all t ∈ E1 (resp. t ∈ E2) is t+ = t + Q2k
(resp. t+= t − Q 2h−1).
Proof. Let t = t0Q0+ t1Q1 + · · · + tnQn be Euler expansion
of t.
1. (a) Suppose that t ∈ F1. It is easily seen that tn > 0 if
n = 2k + 1 and tn < qn− 1 or (tn = qn− 1 and tn−1 = 0)
if n = 2k. Therefore, t+= t + (−1)nQ n.
(b) Suppose that t ∈ F2, then t = t0Q0+ t1Q1+ · · · + t2kQ2k.
Since t < Q2k+1, there exists j ≤ k such that t2j <
q2j. Taking i = max{j | t2j < q2j} we can write t =
t0Q0+ t1Q1+ · · · + t2iQ2i+ [q2i+2Q2i+2+ · · · + q2kQ2k] =
t0Q0+ t1Q1+ · · · + t2iQ2i− Q2i+1+ Q2k+1.
If t2i < q2i− 1 or (t2i = q2i − 1 and t2i−1 = 0) then
t+ = t0Q0+ t1Q1+ · · · + (t2i+ 1)Q2i+ (q2i+1− 1)Q2i+1+
· · · + q2h−1Q2h−1 = t + Q2h− Q2k+1.
If t2i = q2i− 1 and t2i−1 > 0 then t+ = t0Q0 + t1Q1 +
· · · + (t2i−1− 1)Q2i−1+ q2i+1Q2i+1+ · · · + q2h−1Q2h−1 =
2. It is a particular case: Taking qn= 1 we get Pn = Qn−1+Qn =
Qn+1 and Vn = Un. Moreover, in this case we have E1 = F2
and E2 = F1 if n = 2k and E1 = F1and E2 = F2 if n = 2k + 1
2
Let (r = r−1, r0, . . . , rn) and (q0, q1, . . . , qn) be sequences of
positive integers and rn+1 ≥ 0. We suppose that
ri−1 = qiri+ ri+1
(29)
for 0 ≤ i ≤ n.
We thus have r > r0 > r1 > · · · > rn > 0 and rn−1 > rn+1 ≥ 0.
We prove by induction the following identities: r = riQi+1+ ri+1Qi
(30) and
r0Qi = (−1)iri+ rQ1i−1.
(31)
It follows from (31) that
r0Pi+1 = (−1)i(ri − ri+1) + rPi1.
(32)
Let t ∈ IN0. Given t = t0Q0+ t1Q1+ · · · + tnQnits Euler expansion,
we associate with t the following numbers: r(t) = Pn
i=0(−1) it iri and E(t) =Pn i=0tiQ 1
i−1. It follows from (31) that
tr0 = r(t) + rE(t). (33) Moreover, if 0 < t < Qn+1 then rn ≤ r(t) ≤ r − rn (34) and if 0 < t < Pn then r2k ≤ r(t) ≤ r − r2h−1 (35)
Indeed, for 0 < t < Qn+1, let t = t0Q0 + · · · + tnQn be Euler
expansion of t. We can write t = Pk
i=et2iQ2i +
Ph−1
i=e t2i+1Q2i+1
with t2e > 0. Hence, r(t) = Pki=et2ir2i−Ph−1i=e t2i+1r2i+1.
Using r2e =
Ph−1
i=e q2i+1r2i+1+ r2h we get
r(t) = (t2e− 1)r2e+
Pk
i=e+1t2ir2i
+Ph−1
i=e(q2i+1− t2i+1)r2i+1+ r2h.
Now if n = 2k + 1 we get r2h+ r2k+1 ≤ r(t) ≤ k X i=0 q2ir2i= r − r2k+1= r − rn. (36)
If n = 2k there exists j ≤ k such that t2j < q2j since otherwise
t = Qn+1. We then get
rn≤ r(e) ≤ r(t) ≤ r − r2j− r2k+1 ≤ r − r2k− r2k+1.
(37)
Hence, assumption (34) follows. In the particular case that qn = 1
we have Pn = Qn+1 and rn−1 = rn+ rn+1. In this case if n = 2k
we get rn−1 ≤ r(t) ≤ r − rn by (36) and if n = 2k + 1 we get
rn≤ r(t) ≤ r − rn−1 by (37) thus assumption (35) follows 2
Proposition 2. We suppose that rn+1= 0. We equip {0, 1, . . . , r−
1} with the natural order ≤ and Un = {0, 1, . . . , Qn+1− 1} with
Eu-ler order ≤e. Then, the mapping (Un, ≤e) → ({0, 1, . . . , r − 1}, ≤
), t 7→ r(t) is strictly increasing. Proof. Given t ≤e x in Un we have
(t0, −t1, . . . , (−1)jtj, . . . , (−1)ntn) ≤l (x0, −x1, . . . , (−1)jxj, . . . , (−1)nxn)
and r(x)−r(t) = (−1)j(x
j−tj)rj+Pni=j+1(−1)ixiri−Pni=j+1(−1)itiri
1. When j = 2i we get x2i> t2i≥ 0 and x2i+1≤ q2i+1−1. Then,
r(x) − r(t) = (x2i− t2i)r2i+ [−x2i+1r2i+1+ · · · + (−1)nxnrn] −
[−t2i+1r2i+1 + · · · + (−1)ntnrn] ≥ (x2i − t2i)r2i − [(q2i+1 −
1)r2i+1+ · · · + q2h−1r2h−1] − [q2i+2r2i+2+ · · · + q2kr2k] ≥ (x2i−
t2i)r2i− r2i+ r2h+ r2k+1> 0.
2. When j = 2i − 1 we get t2i−1 > x2i−1 ≥ 0 and t2i ≤ q2i−
1. Then, r(x) − r(t) = (t2i−1 − x2i−1)r2i−1+ [x2ir2i+ · · · +
(−1)nx
nrn] − [t2ir2i+ · · · + (−)ntnrn] ≥ (t2i−1− x2i−1)r2i−1−
[q2i+1r2i+1+ · · · + q2h−1r2h−1] − [(q2i− 1)r2i+ · · · + q2kr2k] ≥
(t2i−1− x2i−1)r2i−1+ r2h− r2i−1+ r2k+1> 0 2
Now we consider another sequence of positive integers (s = s−1, s0, s1, . . . , sn) and sn+1 ≥ 0 such that si−1 = qisi + si+1 for
0 ≤ i ≤ n.
We also define s(t) =Pn
i=0(−1) it
isi.
Using (31) we prove the following identity
(rs0− sr0)Qi = (−1)i(rsi− sri).
(38) We derive
(rs0− sr0)Pi+1= (−1)i[r(si− si+1) + s(ri+1− ri)]
(39) from (32) and t(rs0− sr0) = rs(t) − sr(t) (40) from (33). Proof of Theorem 4.
First step : Reduction to the case n = 2k. Suppose that n = n(a, b, c) = 2k + 1.
Since g(a, b, c) = g(b, a, c), it suffices to show that n(b, a, c) is even, L(a, b, c) = L(b, a, c) and l(a, b, c) = l(b, a, c).
We write c = b(a − v0) − a(b − u0). We consider two cases:
If p0 = q0 > 1 then we can write a = (a − v0) + v0, a − v0 =
(q0− 1)v0+ v1, b = (b − u0) + u0, b − u0 = (q0− 1)u0+ u1.
There-fore, n(b, a, c) = n(a, b, c) + 1, L(b, a, c) = L(a, b, c) and l(b, a, c) = l(a, b, c).
If p0 = q0 = 1 we get a − v0 = u1 and b − u0 = v1. We therefore get
a = (q1+ 1)(a − v0) + v2 and b = (p1+ 1)(b − u0) + u2. It follows that
n(b, a, c) = n(a, b, c)−1. Furthermore, we observe that if n(a, b, c) > 1 then we have obviously L(a, b, c) = L(b, a, c) and l(a, b, c) = l(b, a, c). If n(a, b, c) = 1 we have L(b, a, c) = (av1, a(b − (q1 +
1)v1)) = (av1, a(u0− q1v1)) = L(a, b, c) and l(b, a, c) = (bu1, bv2)) =
l(a, b, c). In both cases, n(b, a, c) is even. Therefore, the assumption follows.
Moreover, we can write
c = wnwn+1− vnvn+1.
(41)
Since c > 0, we get wn+1 = (pn− qn)vn+ un+1 > 0 thus pn > qn or
(pn= qn and vn+1 = 0). When pn = qn and vn+1 = 0 we then have
c = wnwn+1 = unun+1.
(42)
In the following steps we suppose that n = 2k. Second step: Case pn > qn.
Taking a = r, v0 = r0, b = s and u0 = w0 = s0 it follows from
(39) that cPn+1 = a(wn− vn+1) + b(wn+1 − vn) ∈ S(a, b) because
wn+1> vn and wn> vn+1. Moreover, it follows from (40) and (35)
that s(t) = b(t) and r(t) = a(t). Hence, for all 0 < t < Pn+1,
tc = ab(t) − ba(t) 6∈ S(a, b) by R4. We conclude that Pn+1 = min{t > 0 | tc ∈ S(a, b)}
(43)
and by (19)-(20), that Vn= {0, 1, . . . , Pn+1− 1} is appropriate. Let
x−t < Pn+1, we have 0 < (x−t)c = (b(x)−b(t))a−(a(x)−a(t))b 6∈
S(a, b). Then, b(x) < b(t) if and only if a(x) < a(t) by R4.
Furthermore, it follows from proposition 2 that b(t)+ − b(t) = b(t+) − b(t) = b(t+ − t). Writing Vn = E1 ∪ E2 with the
no-tation of the lemma, Theorem 2 and (22) show that g(a, b, c) = max{aα1+ β1, aα2+ β2} − (a + b) where α1 = b(t+) − b(t) = b(Qn)
for all t ∈ E1, α2 = b(t+) − b(t) = −b(Qn+1) for all t ∈ E2, and
βi = max Ei.c.
It follows from (31) that α1 = wn, and α2 = vn+1. Moreover, we
see that β2 = c(Pn+1− 1), β1 = c(Qn+1− 1) = −cQn+ cPn+1− c =
bvn− awn+ cPn+1− c by (38). We therefore obtain the formula
g(a, b, c) = cPn+1+ max{avn+1, bvn} − (a + b + c)
(44)
which can be written in the form
f (a, b, c) = a(wn− vn+1) + b(wn+1− vn) + max{avn+1, bvn}
(45)
by (39). Finally,
f (a, b, c) = awn+ bwn+1− min{avn+1, bvn}.
(46)
Third step: Case pn= qn and vn+1= 0.
We have cQn+1 = bwn+1 ∈ S(a, b) by (38) and for all 0 < t < Qn+1,
tc = ab(t) − ba(t) 6∈ S(a, b)) by (34). We deduce that Qn+1 = min{t > 0 | tc ∈ S(a, b)}.
(47)
Using (19)-(20) we show by a similar argument that Un = {0, 1, . . . , Qn+1−
1} is appropriate and Un.c is trimmed. Furthermore, since vn+1 = 0
the lemma show that b(t+) − b(t) = w
n for all t ∈ Un . Therefore,
by (21) f (a, b, c) = awn+ cQn+1= awn+ bwn+1.
b = wnQn+1 by (30). Using (42) we deduce that lcm(b, c) = bwn+1
2
Remark. The case that (pn = qn and vn+1 = 0) can be
deduced from Brauer’s formula (16): We put d = gcd(b, c) = gcd(b, w0) = wn, c0 = dc = wn+1 and b0 = db = Qn+1. Using (30)
we get a = vnQn+1 + wn+1Qn > wn+1Qn+1 = b0c0. We thus have
f (a, b0, c0) = g(a, b0, c0) + a + b0+ c0 = b0c0+ a. Hence, (16) show that f (a, b, c) = df (a, b0, c0) = bc0+ da = lcm(b, c) + a gcd(b, c) 2
Proof of Theorem 5. We first prove by induction that polyno-mials Ri satisfy the following properties: For i ≤ n,
cRi = ayi− bxi,
(48)
cRn+1 = ayn+1− b(enxn− xn−1),
(49)
y−1 = yiRi+1(e0, . . . , ei) − yi+1Ri(e0, . . . , ei−1)
(50)
and for j ≤ i,
yj−1 = yiRi−j+1(ej, . . . , ei) − yi+1Ri−j(ej, . . . , ei−1).
(51)
In the particular case that ej = ej+1 = · · · = ei = 2, we get
Ri−j+1(ej, . . . , ei) = (i − j + 2)
(52) and
yj−1− yj = yj− yj+1 = · · · = (yi− yi+1)
(53)
so in this case (51) can be written in the form yj−1 = (i − j + 2)(yi− yi+1) + yi+1.
Now we consider the set
K = {ei | i = 0 or (0 < i < m and ei > 2)} = {e0 = ek0, ek1, . . . , eks−1}
and we set ks= m. We can suppose that 0 = k0 < k1 < · · · < ks =
m. We have y−1 = (e0− 1)y0+ (y0− y1) (55) and for 0 < i ≤ s, yki−1− yki = (eki− 2)yki+ (yki − yki+1). (56)
Furthermore, (54) show that
yk(i−1) = (ki− k(i−1))(yki−1− yki) + yki
(57)
and (53) that for 0 < i < s,
yk(i−1) − yk(i−1)+1 = · · · = yki−1− yki > yki.
(58)
To apply Theorem 4 we set
v−1 = y−1, u0 = y0, v1 = (y0− y1), q0 = (e0− 1)
and for 0 < i < s,
v2i−1 = yki−1− yki, u2i= yki, q2i = (eki− 2), p2i−1 = (ki− k(i−1)).
Furthermore, if em > 2 we set v2s−1 = (ym−1−ym), v2s+1 = ym+1 = 0, p2s−1 = (m−k(s−1)), u2s = ym, q2s = (em−1) and if em = 2 we set v2s−1 = ym, p2s−1 = (m + 1 − k(s−1)) and u2s = ym+1 = 0. We thus get v−1 > u0 > · · · > v2i−1 > u2i > v2i+1> · · ·
and
v2i−1= q2iu2i+ v2i+1; u2i = p2i+1v2i+1+ u2i+2.
Using (11) we get c = ay0−bx0 = xnyn(dn−en)+xnyn+1−ynxn+1 >
0. Then, dn > en or (dn = en and xn+1 = 0).
To prove (12) we consider two cases: Case n = 0.
We can write a = (d0−1)x0+ (x0−x1) and b = (e0−1)y0+ (y0−y1)
with d0−1 > e0−1 or d0−1 = e0−1 and x1 = 0. Then, L(a, b, c) =
(bx0, b(a − (e0 − 1)x0)), l(a, b, c) = (ay0, a(y0 − y1)). We conclude
that f (a, b, c) = ay0 + b(a − (e0 − 1)x0) − min{a(y0− y1), bx0} =
ay1+ b(a − e0x0) + max{a(y0− y1), bx0} by (7).
Case n > 0.
Let r = max{i | ki < n}.
1. Suppose that dn ≥ en > 2. We can write xn−1− xn = (dn−
2)xn+(xn−xn+1) and yn−1−yn= (en−2)yn+(yn−yn+1) with
dn− 2 > en− 2 or dn− 2 = en− 2 and xn+1 = 0. It follows
that the level of (a, b, c) is even, L(a, b, c) = (bxn, b(xn−1 −
(en − 1)xn) and l(a, b, c) = (ayn, a(yn − yn+1)). Therefore,
f (a, b, c) = ayn+1+ b(xn−1− enxn) + max{a(yn− yn+1), bxn}
by (7).
2. Suppose that dn > en = 2. We thus have k(r+1) > n. we
can write ykr = (k(r+1)− kr)(yk(r+1)−1 − yk(r+1)) + yk(r+1) and
xkr = (n − kr)(xn−1− xn) + xn by (57). Moreover, we have
yk(r+1)−1−yk(r+1) = yn−1−yn= yn−yn+1by (53) and ykr−(n−
kr)(yn−1− yn) = yn by (54). Therefore, the level of (a, b, c) is
odd, L = (a(yn− yn+1), ayn) and l = (b(xn−1− xn), bxn). We
deduce that f (a, b, c) = ayn+1+ b(xn−1− 2xn) + max{a(yn−
3. Suppose that dn = en = 2 and xn+1 = 0 then k(r+1) ≥ n +
1. Using (57) we can write ykr = (k(r+1) − kr)(yk(r+1)−1 −
yk(r+1)) + yk(r+1) = (n + 1 − kr)(yn− yn+1) + yn+1 and xkr =
(n + 1 − kr)xn. Hence, the level of (a, b, c) is odd. Since
ykr − (n + 1 − kr)(yn− yn+1) = yn+1, we get L = (a(yn−
yn+1), ayn+1) and l = (bxn, 0). Therefore, f (a, b, c) = ayn+1+
bxn = ayn+1+ max{a(yn− yn+1), bxn} because, by (48), we
have bxn− ayn+ ayn+1= −cRn+ cRn+1 and it is easily seen
that −cRn+ cRn+1 < 0.
Finally we have proved that
f (a, b, c) = ayn+1+ b[xn−1− enxn] + max{a(yn− yn+1), bxn}2
(59)
To prove Theorem 6 we observe, using (48)-(49), that (11) and the following condition yn+1 Rn+1 ≤ c a < yi Ri for all 0 ≤ i ≤ n. (60)
are equivalent. Taking account of (48)-(49) we obtain f (a, b, c) = cRn+1+ ayn− min{ayn+1, cRn}2 (61) 5. Examples 1. A = {31, 44, 462, 674, 402, 932, 1214}. We take a = 31, b = 44. We obtain (462,∧ 674,∧ 402,∧ 932,∧ 1214) = (22, 26, 30, 40, 42)∧ and (462,∨ 674,∨ 402,∨ 932,∨ 1214) = (5, 3, 12, 7, 2).∨
We remove 674, 932, 1214 from A without altering g(A). We consider (a, b, c1, c2) = (31, 44, 462, 402). Applying Theorem
2. A = {57, 83, 367, 543, 605}. We take a = 57, b = 83. We have (367,∧ 543,∧ 605) = (21, 27, 31),∧ ( ∨ 367, ∨ 543, ∨ 605) = (10, 12, 14), B = b(T )∪{b} = {21, 27, 31, 42, 48, 52, 58, 62, 63, 69, 73, 79, 83}, A = a(T )∪{a} = {10, 12, 14, 20, 22, 24, 26, 28, 30, 32, 34, 36, 57}. We obtain g(A) = 1603. 3. A = {a, b, c} = {137, 250, 337}. We have (ˆc, ˇc) = (101, 54), l = 4, B = {53, 101, 154, 202, 250}, A = {25, 54, 79, 108, 137}. Using Theorem 2 we obtain
g(137, 250, 337) = max{g1, g2, g3} = 7537.
Let us compute g(A) by Theorem 5. We get n(a, b, c) = 1, L(a, b, c) = (6576, 7261), l(a, b, c) = (7250, 6250). We obtain g(a, b, c) = 7250 + 7261 − 6250 − 137 − 250 − 337 = 7537.
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